Question

in the adjoining figure, angle ABC = angle angle AND and D is the mid point of BC.Prove that ∆ADB=~ ∆ADC.

Answer 1

Given in ΔABC, ∠ADC = ∠BAC Consider ΔBAC and ΔADC ∠ADC = ∠BAC (Given) ∠C = ∠C (Common angle) ∴ ΔBAC ~ ΔADC (AA similarity criterion)

Answer 2

the farmolou is
$2.36 \frac{5 {y {x}^{?} }^{2} \times \frac{?}{?} }{?}$